Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>If \(\frac{dy}{dx} + \frac{3}{\cos^2 x}y = \frac{1}{\cos^2 x},\ x \in \left(\frac{-\pi}{3}, \frac{\pi}{3}\right)\), and \(y\!\left(\frac{\pi}{4}\right) = \frac{4}{3}\), then \(y\!\left(-\frac{\pi}{4}\right)\) equals:</p>
<p>\(\dfrac{1}{3} + e^6\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(-\dfrac{4}{3}\)</p>
<p>\(\dfrac{1}{3} + e^3\)</p>
Step-by-Step Solution
Key Concept: This is a first-order linear ODE of the form dy/dx + P(x)y = Q(x). Use the integrating factor method with e^∫P(x)dx = e^(3tan x), then apply the initial condition to find the particular solution.
<p><strong>Step 1:</strong> Identify the linear form: dy/dx + (3/cos²x)y = 1/cos²x, where P(x) = 3sec²x and Q(x) = sec²x.</p><p><strong>Step 2:</strong> Find integrating factor: IF = e^(∫3sec²x dx) = e^(3tan x).</p><p><strong>Step 3:</strong> Multiply both sides by e^(3tan x):<br/>e^(3tan x)·dy/dx + 3sec²x·e^(3tan x)·y = sec²x·e^(3tan x)</p><p><strong>Step 4:</strong> Recognize LHS as d/dx[e^(3tan x)·y]:<br/>d/dx[e^(3tan x)·y] = sec²x·e^(3tan x)</p><p><strong>Step 5:</strong> Integrate both sides:<br/>e^(3tan x)·y = ∫sec²x·e^(3tan x) dx</p><p>Let u = 3tan x, then du = 3sec²x dx, so sec²x dx = du/3:<br/>e^(3tan x)·y = (1/3)∫e^u du = (1/3)e^u + C = (1/3)e^(3tan x) + C</p><p><strong>Step 6:</strong> Therefore: y = 1/3 + C·e^(-3tan x)</p><p><strong>Step 7:</strong> Apply initial condition y(π/4) = 4/3:<br/>4/3 = 1/3 + C·e^(-3·1)<br/>1 = C·e^(-3)<br/>C = e^3</p><p><strong>Step 8:</strong> Solution: y = 1/3 + e^3·e^(-3tan x) = 1/3 + e^(3-3tan x)</p><p><strong>Step 9:</strong> Evaluate at x = -π/4:<br/>tan(-π/4) = -1<br/>y(-π/4) = 1/3 + e^(3-3(-1)) = 1/3 + e^6</p><p>∴ Answer: A</p>
Correct Answer: A